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cos-7x-cos-8x-1-2cos-5x-dx-

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Question Number 136992 by liberty last updated on 28/Mar/21
ℓ = ∫ ((cos 7x−cos 8x)/(1+2cos 5x)) dx =?
$$\ell\:=\:\int\:\frac{\mathrm{cos}\:\mathrm{7x}−\mathrm{cos}\:\mathrm{8x}}{\mathrm{1}+\mathrm{2cos}\:\mathrm{5x}}\:\mathrm{dx}\:=? \\ $$
Answered by EDWIN88 last updated on 28/Mar/21
ℓ = ∫ ((cos 7x−cos 8x)/(1+2cos 5x)) dx Simplify the integrand by Euler′s formula ((cos 7x−cos 8x)/(1+2cos 5x)) = ((e^(i7x) +e^(−i7x) −e^(i8x) −e^(−i8x) )/(2(1+e^(i5x) +e^(−i5x) ))) making substitution e^(ix) = r , this becomes ((r^7 +r^(−7) −r^8 −r^(−8) )/(2(1+r^5 +r^(−5) ))) = − ((r^(16) −r^(15) −r+1)/(2r^3 (1+r^5 +r^(10) ))) = − (((r−1)(r^(15) −1))/(2r^3 (1+r^5 +r^(10) )))= − (((r−1)(r^5 −1))/(2r^3 )) = − ((r^6 −r^5 −r+1)/(2r^3 )) = − (((r^6 +1)−(r^5 −r))/(2r^3 )) = ((r^2 −r^(−2) )/2) − ((r^3 −r^(−3) )/2) = cos 2x−cos 3x So the integral becomes ℓ=∫ (cos 2x−cos 3x)dx ℓ = ((sin 2x)/2) − ((sin 3x)/3) + C
$$\:\ell\:=\:\int\:\frac{\mathrm{cos}\:\mathrm{7x}−\mathrm{cos}\:\mathrm{8x}}{\mathrm{1}+\mathrm{2cos}\:\mathrm{5x}}\:\mathrm{dx} \\ $$$$\mathrm{Simplify}\:\mathrm{the}\:\mathrm{integrand}\:\mathrm{by}\:\mathrm{Euler}'\mathrm{s}\:\mathrm{formula} \\ $$$$\frac{\mathrm{cos}\:\mathrm{7x}−\mathrm{cos}\:\mathrm{8x}}{\mathrm{1}+\mathrm{2cos}\:\mathrm{5x}}\:=\:\frac{\mathrm{e}^{\mathrm{i7x}} +\mathrm{e}^{−\mathrm{i7x}} −\mathrm{e}^{\mathrm{i8x}} −\mathrm{e}^{−\mathrm{i8x}} }{\mathrm{2}\left(\mathrm{1}+\mathrm{e}^{\mathrm{i5x}} +\mathrm{e}^{−\mathrm{i5x}} \right)} \\ $$$$\mathrm{making}\:\mathrm{substitution}\:\mathrm{e}^{\mathrm{ix}} \:=\:\mathrm{r}\:,\:\mathrm{this}\:\mathrm{becomes} \\ $$$$\frac{\mathrm{r}^{\mathrm{7}} +\mathrm{r}^{−\mathrm{7}} −\mathrm{r}^{\mathrm{8}} −\mathrm{r}^{−\mathrm{8}} }{\mathrm{2}\left(\mathrm{1}+\mathrm{r}^{\mathrm{5}} +\mathrm{r}^{−\mathrm{5}} \right)}\:=\:−\:\frac{\mathrm{r}^{\mathrm{16}} −\mathrm{r}^{\mathrm{15}} −\mathrm{r}+\mathrm{1}}{\mathrm{2r}^{\mathrm{3}} \left(\mathrm{1}+\mathrm{r}^{\mathrm{5}} +\mathrm{r}^{\mathrm{10}} \right)} \\ $$$$=\:−\:\frac{\left(\mathrm{r}−\mathrm{1}\right)\left(\mathrm{r}^{\mathrm{15}} −\mathrm{1}\right)}{\mathrm{2r}^{\mathrm{3}} \left(\mathrm{1}+\mathrm{r}^{\mathrm{5}} +\mathrm{r}^{\mathrm{10}} \right)}=\:−\:\frac{\left(\mathrm{r}−\mathrm{1}\right)\left(\mathrm{r}^{\mathrm{5}} −\mathrm{1}\right)}{\mathrm{2r}^{\mathrm{3}} } \\ $$$$=\:−\:\frac{\mathrm{r}^{\mathrm{6}} −\mathrm{r}^{\mathrm{5}} −\mathrm{r}+\mathrm{1}}{\mathrm{2r}^{\mathrm{3}} }\:=\:−\:\frac{\left(\mathrm{r}^{\mathrm{6}} +\mathrm{1}\right)−\left(\mathrm{r}^{\mathrm{5}} −\mathrm{r}\right)}{\mathrm{2r}^{\mathrm{3}} } \\ $$$$=\:\frac{\mathrm{r}^{\mathrm{2}} −\mathrm{r}^{−\mathrm{2}} }{\mathrm{2}}\:−\:\frac{\mathrm{r}^{\mathrm{3}} −\mathrm{r}^{−\mathrm{3}} }{\mathrm{2}}\:=\:\mathrm{cos}\:\mathrm{2x}−\mathrm{cos}\:\mathrm{3x} \\ $$$$\mathrm{So}\:\mathrm{the}\:\mathrm{integral}\:\mathrm{becomes}\:\ell=\int\:\left(\mathrm{cos}\:\mathrm{2x}−\mathrm{cos}\:\mathrm{3x}\right)\mathrm{dx} \\ $$$$\ell\:=\:\frac{\mathrm{sin}\:\mathrm{2x}}{\mathrm{2}}\:−\:\frac{\mathrm{sin}\:\mathrm{3x}}{\mathrm{3}}\:+\:\mathrm{C} \\ $$
Answered by som(math1967) last updated on 29/Mar/21
∫ ((sin5x2sin((15x)/2)sin(x/2))/(sin5x+2sin5xcos5x))dx ∫((2sin5xsin((15x)/2)sin(x/2))/(sin5x+sin10x))dx ∫((4sin((5x)/2)cos((5x)/2)sin((15x)/2)sin(x/2))/(2sin((15x)/2)cos((5x)/2)))dx ∫2sin((5x)/2)sin(x/2)dx ∫cos2x−cos3xdx (1/2)sin2x−(1/3)sin3x +C
$$\int\:\frac{{sin}\mathrm{5}{x}\mathrm{2}{sin}\frac{\mathrm{15}{x}}{\mathrm{2}}{sin}\frac{{x}}{\mathrm{2}}}{{sin}\mathrm{5}{x}+\mathrm{2}{sin}\mathrm{5}{xcos}\mathrm{5}{x}}{dx}\: \\ $$$$\int\frac{\mathrm{2}{sin}\mathrm{5}{xsin}\frac{\mathrm{15}{x}}{\mathrm{2}}{sin}\frac{{x}}{\mathrm{2}}}{{sin}\mathrm{5}{x}+{sin}\mathrm{10}{x}}{dx} \\ $$$$\int\frac{\mathrm{4}{sin}\frac{\mathrm{5}{x}}{\mathrm{2}}{cos}\frac{\mathrm{5}{x}}{\mathrm{2}}{sin}\frac{\mathrm{15}{x}}{\mathrm{2}}{sin}\frac{{x}}{\mathrm{2}}}{\mathrm{2}{sin}\frac{\mathrm{15}{x}}{\mathrm{2}}{cos}\frac{\mathrm{5}{x}}{\mathrm{2}}}{dx} \\ $$$$\int\mathrm{2}{sin}\frac{\mathrm{5}{x}}{\mathrm{2}}{sin}\frac{{x}}{\mathrm{2}}{dx} \\ $$$$\int{cos}\mathrm{2}{x}−{cos}\mathrm{3}{xdx} \\ $$$$\:\frac{\mathrm{1}}{\mathrm{2}}{sin}\mathrm{2}{x}−\frac{\mathrm{1}}{\mathrm{3}}{sin}\mathrm{3}{x}\:+{C} \\ $$$$ \\ $$

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